SDSU

 

Math 536 -Mathematical Modeling
Fall Semester, 2000
Linear Discrete Dynamical Models - Examples
 © 1999, All Rights Reserved, SDSU & Joseph M. Mahaffy
San Diego State University -- This page last updated 31-Mar-00

Linear Discrete Dynamical Models - Worked Examples

This section is still under development.

Example 1: For each of the following linear discrete dynamical systems, find the first three iterations, y1, y2, and y3. Also, determine the equilibrium value and determine if it is stable or not.

a. yn+1 = 1.05 yn - 200 with y0= 2000.

b. yn+1 = 0.6 yn + 50 with y0= 100.

Solutions:

a. For y1, n = 0. Substituting the given value y0= 2000:

y1 = 1.05y0 - 200 = 1.05(2000) - 200 = 1900

Using the value for y1 we can find y2

y2 = 1.05y1 - 200 = 1.05(1900) - 200 = 1795

y3 = 1.05y2 - 200 = 1.05(1795) - 200 = 1684.75

To find the equilibrium value replace both yn+1 and yn with ye. This is because at equilibrium, all iterations yield exactly the same results.

Thus, ye = 1.05ye - 200, or 0.05ye = 200, and so ye = 4000. From above, we can see that as n increases, the value of yn moves away from the equilibrium point, ye. Note that the value y0 < ye, with yn continually decreasing. The slope of the line is 1.05 on the right hand side, which is greater than 1. This is characteristic of unstable linear discrete dynamical models.

b. For y1, n = 0. Substituting the given value y0= 100 :

y1 = 0.6 y0 + 50 = 0.6(100) + 50 = 110

y2 = 0.6 y1 + 50 = 0.6(110) + 50 = 116

y3 = 0.6 y2 + 50 = 0.6(116) + 50 = 119.6

For the equilibrium, we again let ye = yn+1 = yn, so that ye = 0.6ye + 50. Thus, 0.4y = 50, or y = 125. In this case, the solution for y is increasing towards the equilibrium, so that the equilibrium is stable. Note that for this case the slope of the linear model, 0.6, is less than 1, which is characteristic of stable linear discrete dynamical models.
 
 

Example 2: A subject with an unknown lung ailment enters the lab for testing. She is given a supply of air that has an enriched amount of argon gas (Ar). (Recall that atmospheric argon occurs at 0.93% or a concentration of g = 0.0093.) After breathing this supply of enriched gas, two successive breaths are measured with c1 = 0.0736 and c2 = 0.0678 of Ar. The model for breathing is given by

cn+1 = (1 - q) cn + q g.

Find the fraction of air breathed, q. What is the concentration of argon remaining in her lungs after 5 breaths?

Assume that her tidal volume is measured to be Vi = 220. Find the functional reserve volume, Vr, where q = Vi/(Vi + Vr).

Solution:

Since we are given g and two consecutive values of cn+1, we can find q as shown below, using

c2 = (1 - q)c1 + gq

0.0678 = (1 - q)(0.0736) + 0.0093q

q = 0.0902

To find the concentration of Ar in her lungs after 5 breaths, we need to know what c5 is.

c2 = (1 - q)c1 + gq = 0.0678 as given above

c3 = (1 - 0.0902)c2 + 0.0093(0.0902) = 0.9098(0.0678) + 0.000839 = 0.06252

c4 = 0.9098(0.06252) + 0.000839 = 0.05772

c5 = 0.9098(0.05772) + 0.000839 = 0.05336

To find the functional reserve volume we use the relationship q = Vi/(Vi + Vr).

Thus, the tidal volume Vr = 2219.
 
 

Example 3: A population of animals in a particular lake grows according to the Mathusian growth law. In addition, a constant number are entering the lake from a river. Thus, this population satisfies the discrete Malthusian growth model with immigration given by the equation:

Pn+1 = (1 + r)Pn + m,

where r is the rate of growth and m is the constant number entering the lake. In three successive weeks, the population is measured at P0 = 500, P1 = 670, and P2 = 874. Find the rate of growth r and immigration rate m, then determine the populations expected in the next two weeks.

Solution:

Substituting the given information into the discrete Malthusian growth model gives 2 equations and 2 unknowns (r and m).

P1 = (1 + r)P0 + m and P2 = (1 + r)P1 + m

670 = (1 + r)500 + m and 874 = (1 + r)670 + m

Next, we solve for r using substitution for m.

670 = (1 + r)(500 - 670) + 874

1 + r = 1.2, or r = 0.2

Substituting this value for the rate of growth r back into either equation and solving for the immigration rate m, we find that m = 70. Thus, the model can be rewritten as

Pn+1 = 1.2Pn + 70

With this model we can determine the populations expected in the next two weeks, P3 and P4.

P3 = 1.2P2 + 70 = 1.2(874) + 70 = 1118.8

P4 = 1.2(1118.8) + 70 = 1412.56